A small coil C with N = 200 turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in figure. The cross sectional area of coil is A=1.0 cm2, length of arm OA of the balance beam is l = 30 cm. When there is no current in the coil the balance is in equilibrium. On passing a current I = 22 mA through the coil the equilibrium is restored by putting the additional counter weight of mass ∆m=60 mg on the balance pan. Find the magnetic induction at the spot where coil is located.
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a
0.4 T
b
0.3 T
c
0.2 T
d
0.1 T
answer is A.
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Detailed Solution
On passing current through the coil, it acts as a magnetic dipole. Torque acting on magnetic dipole is counter balanced by the moment of additional weight about position O. Torque acting on a magnetic dipoleτ=MB sin θ=NiAB sin 90o=NiABAgain, τ = Force x Lever arm = ∆mg x l⇒NiAB = ∆mgl⇒B=∆mglNiA=60 x 10-3 x 9.8 x 30 x 10-2200 x 22 x 10-3 x 1 x 10-4=0.4 T